How would you solve this trig function:
cos2x=sin^2-2
how do you know what trig identities to use?
cos 2x = cos^2 x - sin^2 x identity
so we really have
cos^2 x - sin^2 x = sin^2 x - 2
cos^2 x = 2 (sin^2 x-1)
but sin^2 x = 1 - cos^2 x
so we have
cos^2 x = 2 (-cos^2 x)
or
3 cos^2 x = 0
well cos of Pi/2 (90deg) or 3 pi/2 (270 deg) = 0
1 - 2sin^2x = sin^2x -2
3 sin^2x = 3
sin^2x = 1
sinx = +1 or -1
x = pi/2 or 3 pi/2
oh i get it.... how do you know what identities to use? did you memorize them?
The only ones I have memorized are
cos 2x = cos^2x - sin^2 x
sin 2x = 2 sinx cosx,
and cos^x + sin^2 x = 1.
I have an old book with many others that I keep handy.
To solve the given equation cos2x = sin^2(x) - 2, let's break down the process step by step.
First, we need to determine which trigonometric identities might be useful. Generally, it's a good idea to simplify the equation using basic trigonometric identities such as the Pythagorean identity, sum and difference identities, and double-angle identities.
In this case, we can use the double-angle identity for cosine, which states that cos(2θ) = cos^2(θ) - sin^2(θ). This identity will help us simplify the given equation.
Now, let's substitute cos(2x) with its double-angle identity:
cos(2x) = cos^2(x) - sin^2(x) - 2
The equation now becomes:
cos^2(x) - sin^2(x) - 2 = sin^2(x) - 2
Next, let's collect like terms and move everything to one side of the equation:
cos^2(x) - sin^2(x) - sin^2(x) + 2 = 0
Simplifying further:
cos^2(x) - 2sin^2(x) + 2 = 0
We now have a quadratic-like equation in terms of sine and cosine.
To proceed further, we can use a trigonometric identity known as the Pythagorean identity:
sin^2(x) + cos^2(x) = 1
Rearranging this identity, we get:
sin^2(x) = 1 - cos^2(x)
Substituting this into the equation:
1 - cos^2(x) - 2cos^2(x) + 2 = 0
Simplifying:
-3cos^2(x) + 3 = 0
Now, divide the equation by -3:
cos^2(x) - 1 = 0
Factorizing the equation:
(cos(x) - 1)(cos(x) + 1) = 0
Setting each factor to zero:
cos(x) - 1 = 0 or cos(x) + 1 = 0
Solving for cos(x):
cos(x) = 1 or cos(x) = -1
Now, to find the values of x, we need to think about the domain of the cosine function.
The cosine function has a range between -1 and 1, inclusive. Therefore, the only valid solution is when cos(x) = -1.
Solving cos(x) = -1, we find x = π + kπ, where k is an integer.
So, the solution to the equation cos2x = sin^2-2 is x = π/2 + kπ, where k is an integer.